Problem: Find $\dfrac{d}{dx}\log_2(2x^2-7x+1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\ln(2)}{4x-7} $ (Choice B) B $\dfrac{1}{(2x^2-7x+1)\ln(2)}$ (Choice C) C $\dfrac{4x-7}{(2x^2-7x+1)\ln(2)}$ (Choice D) D $\dfrac{\ln(2)(4x-7)}{(2x^2-7x+1)}$
Explanation: $\log_2(2x^2-7x+1)$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=2x^2-7x+1$, then $\log_2(2x^2-7x+1)=\log_2\Bigl(u(x)\Bigr)$. $\dfrac{d}{dx}\log_2(2x^2-7x+1)$ can be found using the following identity: $\dfrac{d}{dx}\left[\log_2\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{\ln(2)u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\log_2(2x^2-7x+1) \\\\ &=\dfrac{d}{dx}\log_2\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=2x^2-7x+1} \\\\ &=\dfrac{u'(x)}{\ln(2)u(x)} \\\\ &=\dfrac{4x-7}{\ln(2)(2x^2-7x+1)}&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=\dfrac{4x-7}{(2x^2-7x+1)\ln(2)} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\log_2(2x^2-7x+1)=\dfrac{4x-7}{(2x^2-7x+1)\ln(2)}$.